QUESTION 7 7.1 Determine \( f^{\prime}(x) \) from first principles if \( f(x)=\frac{1}{x} \). 7.2 Determine: 7.2.1 \( \frac{d}{d x}\left(\sqrt{4 x^{6}}+\sqrt{2} \cdot x^{2}\right) \) 7.2.2 \( g^{\prime}(x) \) if \( g(x)=\frac{3 x^{4}-4 x^{2}+6}{x^{2}} \) (3) 7.3 The equation of the tangent to \( f(x)=3 x^{2}+b x+c \) at \( x=1 \) is given by \( y=9 x-9 \). Determine the values of \( b \) and \( c \). (4) [15]

Below is the full solution with step‐by‐step reasoning. ───────────────────────────── Problem 7.1 ───────────────────────────── We wish to find f′(x) from first principles for   f(x) = 1/x. Using the definition of the derivative:   f′(x) = lim₍ₕ→0₎ [f(x + h) − f(x)] / h Step 1. Write the difference quotient:   f′(x) = lim₍ₕ→0₎ [(1/(x + h) − 1/x) / h]. Step 2. Combine the fractions inside the bracket:   1/(x+h) − 1/x = (x − (x+h)) / [x(x+h)] = −h / [x(x+h)]. Thus,   f′(x) = lim₍ₕ→0₎ [ (−h / (x(x+h))) / h ]. Step 3. Cancel h (as long as h ≠ 0):   f′(x) = lim₍ₕ→0₎ [−1 / (x(x+h))]. Step 4. Take the limit as h → 0:   f′(x) = −1 / (x·x) = −1/x². ───────────────────────────── Problem 7.2.1 ───────────────────────────── Determine the derivative of   f(x) = √(4x⁶) + √2 · x². Step 1. Rewrite √(4x⁶):   √(4x⁶) = √4 · √(x⁶) = 2|x³|. For many problems, the domain is assumed to be x ≥ 0 so that |x³| = x³. (If x can be negative, you must account for the absolute value; here we assume x ≥ 0.) Thus, assume x ≥ 0 and get:   f(x) = 2x³ + √2 · x². Step 2. Differentiate term-by-term:   d/dx (2x³) = 6x²,   d/dx (√2 · x²) = 2√2 x. The derivative is:   f′(x) = 6x² + 2√2 x. ───────────────────────────── Problem 7.2.2 ───────────────────────────── Find g′(x) for   g(x) = (3x⁴ − 4x² + 6) / x². Step 1. Divide each term in the numerator by x²:   g(x) = 3x² − 4 + 6/x². Step 2. Differentiate term-by-term:   d/dx (3x²) = 6x,   d/dx (−4) = 0,   d/dx (6/x²) = 6 · d/dx (x^(−2)) = 6(−2)x^(−3) = −12/x³. Thus,   g′(x) = 6x − 12/x³. ───────────────────────────── Problem 7.3 ───────────────────────────── The function is   f(x) = 3x² + bx + c, and the equation of the tangent at x = 1 is given as   y = 9x − 9. The tangent line must satisfy two conditions at x = 1: 1. It must pass through the point (1, f(1)). 2. Its slope must equal f′(1). Step 1: Use the point of tangency.   f(1) = 3(1)² + b(1) + c = 3 + b + c. The tangent line at x = 1 gives y = 9(1) − 9 = 0. Thus,   3 + b + c = 0  →  b + c = −3.                  (Equation 1) Step 2: Use the slope.   f′(x) = derivative of 3x² + bx + c = 6x + b. Then, at x = 1:   f′(1) = 6 + b. Since the slope of the tangent is 9, we have:   6 + b = 9  →  b = 3. Step 3: Substitute b = 3 into Equation 1:   3 + c = −3  →  c = −6. ───────────────────────────── Final Answers Summary ───────────────────────────── 7.1   f′(x) = −1/x². 7.2.1  d/dx [√(4x⁶) + √2·x²] = 6x² + 2√2 x      (for x ≥ 0). 7.2.2  g′(x) = 6x − 12/x³. 7.3   For f(x) = 3x² + bx + c with tangent y = 9x − 9 at x = 1, we have b = 3 and c = −6.